solution for midterm exam Ch 01 – Ch 06

1. Calculate each value requested for the following set of scores (2 points each) X: 1, 3, 0, 2 and Y: 5, 1, –2, –4 a. ΣX = 6 b. ΣY = 0 c. ΣXY = 0 d. ΣX2 = 1+9+0+4= 14 e. Σ(Y – 2)2 = 9+1+16+36= 62

2.For the distribution shown in the following table: (3 points each) X f cf c% 25-29 4 25 100 20-24 6 21 84 15-19 7 15 60 10-14 5 8 32 5-9 3 12 Find the percentile rank for X = 14.5. (32%) Find the 60th percentile. (19.5) Find the percentile rank for X = 11. (18%) Find the 66th percentile. (20.75)

2. c-d c. (11-9.5)/(14.5-9.5) = (x-0.12)/(0.32-0.12) 1.5/5=(x-0.12)/0.2 0.3*0.2=x-0.12 0.06+0.12=x x = 0.18 d. (y-19.5)/(24.5-19.5)=(0.66-0.6)/(0.84-0.6) (y-19.5)/5=0.06/0.24y-19.5=5*0.25=1.25 y = 19.5+1.25 = 20.75

3. (5 points each) A set of n = 6 scores has a mean of M = 10. Another set of scores has n = 4 and M = 15. If these two sets of scores are combined, what is the mean for the combined group? (n=10, ΣX = 120, M = 12) A sample of n = 7 scores has a mean of M = 6. If one score with a value of X = 12 is removed from the sample, what is the mean for the remaining scores? (ΣX = 42 ΣX’ = 42-12=30M’=30/6=5) A sample of n = 6 scores has a mean of M = 10. One new score is added to the sample and the new mean is computed to be M = 9. What is the value of the score that was added to the sample? (ΣX = 60 ΣX’ =60+x=9*7, X=3) For a sample with M = 40 and s = 4, the middle 95% of the individuals will have scores between what scores? M2s=(32,48)

4. (5 points each) A population has µ = 50 and σ = 5. If 10 points are added to every score in the population, then what are the new values for the mean and standard deviation? (µ = 50+10 and σ = 5) A population of scores has µ = 50 and σ = 5. If every score in the population is multiplied by 3, then what are the new values for the mean and standard deviation? (µ = 50*3 and σ = 5*3) Using the definitional formula, compute SS, variance and the standard deviation for the following sample of scores. Scores: 3, 6, 1, 6, 5, 3 (M=4, SS = 20; s 2 = 4; s = 2)

5. (9 points each) A population of scores with µ = 73 and σ = 20 is standardized to create a new population with µ = 50 and σ = 10. What is the new value for each of the following scores from the original population? Scores: 63, 65, 77, 83 For a distribution of scores, X = 40 corresponds to a z‑score of z = +1.00, and X = 28 corresponds to a z‑score of z = –0.50. What are the values for the mean and standard deviation for the distribution? (Hint: Sketch a distribution and locate each of the z‑score positions.)

5. X=63z=(63-73)/20=−0.5(X’-50)/10= − 0.5 X’=45 b. (40 − µ)/ σ = 1  µ + σ = 40 (28 − µ)/ σ = −0.5  µ − 0.5 σ = 28 1.5 σ = 40-28 = 12  σ =8  µ = 40-8 = 32

6. A normal distribution has a mean of µ = 100 with σ = 20 6. A normal distribution has a mean of µ = 100 with σ = 20. Find the following probabilities: (3 points each) use z table a. p(X > 102)=p(z>102-100/20) =p(z>0.1)=0.5-0.0398=0.4602 b. p(X < 65)=p(z<65-100/20)=p(z<-1.75)=0.5-0.4599=0.0401 c. p(X < 130)= p(z<130-100/20)=p(z<1.5)=0.5+0.4332=0.9332 d. p(95 < X < 105)= p(95-100/20<z<105-100/20)=p(-0.25<z<0.25) =2*0.0987 = 0.1974 e. What z-score separates the highest 30% from the rest of the scores? p(0<z<z0)=0.1985≈0.2z0=0.52

7. In an ESP experiment subjects must predict whether a number randomly generated by a computer will be odd or even. (hint: probability of odd or even is the same as the probability of head or tail by tossing a fair coin) (5 points each) a. What is the probability that a subject would guess exactly 18 correct in a series of 36 trials? With n = 36 and p = q = 1/2, you may use the normal approximation with µ = np = 18 and  2 = npq = 9   = 3. X = 18 has real limits of 17.5 and 18.5 corresponding to z = 0.17 and z = +0.17. p(x=18) = p (-0.17<z<0.17)=2*p(0<z<0.17)=2*0.0675= 0.135

7b. b. What is the probability that a subject would guess more than 20 correct in a series of 36 trials? X > 20  X > 20.5  z > 20.5-18/3=0.83 p(X>20.5) = p(z>0.83) = 0.5-p(0<z<0.83) = 0.5 - 0.2967 = 0.2033